Question #bf123

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Aug 24, 2017

Answer:

i) 2 loops
ii) 16 loops

Explanation:

The equation relating tension and speed of wave is given by:

#v=sqrt(T/mu)#
where: #T" is the tension"; mu" is the linear density of the string"#

Since we know that #v=flambda#, we can rewrite the equation into:

#flambda=sqrt(T/mu)#

I)Tension increased by a factor of 4

Let #T# be the initial tension of the string
#:.4T# will be the final tension of the string

#4#loops are formed,
#L=2lambda#

The string used is the same, thus the linear density, #mu# is a constant.
The frequency of the vibrator is also kept as the same, #f# is another constant.

Rearranging
#flambda=sqrt(T/mu)#

#f^2mu=T/lambda^2#

since #f# and #mu# are constants, their product is a constant too.

#T_i/lambda_i^2=T_f/lambda_f^2#
subscript #i# for initial, #f# for final

#T/(L/2)^2=(4T)/lambda_f^2#

#(4T)/(4(L^2/4))=(4T)/lambda_f^2#

#4(L^2/4)=lambda_f^2#

#lambda_f^2=L^2#

#lambda_f=L#

Therefore the number of loops formed will be 2.

II)Frequency increased by a factor of 4

Now, Tension instead of frequency becomes constant.

#flambda=sqrt(T/mu)#

Since Tension and Linear density remain the same, #sqrt(T/mu)# is constant.

Thus,
#f_ilambda_i=f_flambda_f#

Let #f# be the initial frequency,
#4f# will be the final frequency.

#4#loops are formed,
#L=2lambda#
#lambda=L/2#

#f(L/2)=(4f)lambda_f#

#lambda_f=f(L/2)/(4f)#

#lambda_f=L/8#

Thefore, the number of loops is 16.

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